Question

I need a solution for this advanced mechanics of steel structure  problem:

A composite beam of rectangular cross section is constructed by joining material “A” with material “B": material “A” is joined to material “B” in such a manner so that fully composite behavior is achieved, i.e plane sections remain plane over the entire cross section. Material "A" Material “B” Given: Material "A" and "B" are both elastic -perfectly plastic, with the following properties: Material "A": E = 30,000 ksi and o=50ksi Material "B": E = 10,000 ksi and Oy=20ksi Bending moment is applied about the section's strong axis. Compute a. For bending in the elastic range, compute the effective EI. That is, what is the slope of the moment curvature (M-o) diagram in the elastic region. b. Compute Moment (M) when first yield occurs at the bottom fiber. c. Compute Moment (M) when first yield occurs at the top fiber. At this moment (M) how much of material “B” has yielded? d. Compute the fully plastic moment.

Answer 1

Hello!!

The answer is given below.

- EA= 30000ksi Oy A- soksi EB= 100ooksi JyB= 20ksi Sin & wint a) Effective E = E=EAAA + ER TAB AA+ AB 3000OX 4X5 + 1000Oxu x5 4x10 2 20000lesi Effective I Te = bd? _ 4x03 - 1000 = 333.33zint Effective F1 => 20000X1000 = 2x10' Kip-in2

6) Composite section Plane remains plane Top te Bottom fibre yields when extreme = Sye zolesi TI Se Sin -12 I sin 2oksi » Moment M= 20x 1 x SX 2/₂ x5 = 166. 67 kip-in = 13.889 kip ft c) Top fibre yields when extreme - TJA soksi yl-5-20x5 y'azin Walin zoksi Moment - 50x0.5x5x12x 5 + 20x 12 x 2 x 1/2 72 + 20x3x2.5 = 371.667 kip-in = 30.972 kip-fr. Material Balthas yielded for a extrane fibre. depth of sin from bottom

go ksi 20 legi Full plasticity is reached when both the materials yield for the complete depth. Fully plastic moment => Mp = 20X5x 2.5+ 50x5x25 = 875 kip in My = 72.9167 kip-ft